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fredd [130]
3 years ago
12

Which of the following pieces of laboratory equipment is not directly used to make measurements

Physics
2 answers:
pychu [463]3 years ago
7 0

Answer:

A. Test tube.

Explanation:

The test tubes are used to collect samples of a particular liquid.

A graduated cylinder is like a test tube but it has marks on the tube to measure the volume of a liquid.

A ruler is used to measure length.

And finally, a thermometer is used to measure the temperature.

I hope this answer helps you.

RoseWind [281]3 years ago
4 0
I believe the answer is a test tube.
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A 2 kg ball is thrown upward with a velocity of 15 m/s. What is the kinetic energy of the ball as it is being thrown?
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A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o
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3 years ago
3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a
sasho [114]

Explanation:

Given that,

Number of turns in the coil, N = 100

Area of the coil, A = 100 cm² = 0.01 m²

It is placed at an angle of 70°.

Magnetic field, B = 0.1 Wb/m²

We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.

Magnetic flux is given by :

\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb

So, the magnetic flux through the coil is 0.1 Wb.

Emf induced in the coil is :

\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{0.034}{10^{-3}}\\\\=34\ V

So, 34V of emf is induced in the coil.

7 0
3 years ago
The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co
Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density = 8.4\times10^{28}\ m^3

Current i = 4.6\times10^{18}\ electron/s

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

v_{d}=\dfrac{I}{neA}

v_{d}=\dfrac{Ne}{tne\times\pi r^2}

Put the value into the formula

v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}

v_{d}=4.842\times10^{-5}\ m/s

We need to calculate the time

Using formula for time

v_{d}=\dfrac{l}{t}

t=\dfrac{l}{v_{d}}

Where, l = length

v_{d} = drift velocity

Put the value into the formula

t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}

t=6402.31\ sec

t=106.7\ minute

Hence, The time is 106.7 minute.

7 0
3 years ago
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